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Marat540 [252]
3 years ago
15

MULTIPLE CHOICE

Physics
1 answer:
almond37 [142]3 years ago
7 0

Answer:

1st law known as low of inertial

2nd law of acelaration and force relationship

3rd law of action and reaction

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A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d
QveST [7]

Answer:

  The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

  (target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

  (10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

  (target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

  (target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

  -94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

  13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

  23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

  (12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

  target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

  √(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0
3 years ago
When object goes under acceleration
lidiya [134]

when object goes under acceleration

c).its velocity always increases

<h3><u>Additional</u><u> </u><u>information</u><u>:</u><u>-</u></h3>

★ Acceleration: Rate of increase in velocity.

★ Velocity: Distance travelled by a body per unit time in given direction is called velocity .

6 0
3 years ago
Read 2 more answers
9. Pick the best example of Newton's Second Law in action.
madreJ [45]

Answer:

A rocket taking off from earth which pushes gasses in one direction and the rocket in

the other

7 0
3 years ago
What wave on the electromagnetic spectrum has the highest frequency?
Jet001 [13]

Answer:

Explanation:

Answer: Gamma rays

Gamma rays have the highest frequency.

What is an electromagnetic wave?

An electromagnetic wave requires no medium for its propagation.

It consists of a spectrum of different wavelengths.

Different wavelengths of rays have different energies and different frequencies.

Higher frequency rays have the highest energies.

What is gamma-ray?

These are ionized radiations.

Gamma radiations are obtained from the decay of the atomic nucleus.

It has the highest frequency which is why it can penetrate through matter.

It has the smallest wavelength and highest energy.

The frequency of gamma rays is more than 10^19 cycles per second and wavelength less than 100 picometers.

6 0
2 years ago
Read 2 more answers
A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the 14) car a deceleration of 3.50
sergij07 [2.7K]
To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving. 

First, we manipulate the one of the kinematic equations
 
v^2 = v0^2 + 2 (a) (x)  where v = 0 since the car stopped

Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,

<span>we get v0 = sqrt (2(a)(x))

Substituting the known values,

v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s 
</span>
Therefore, before stopping the car the original speed of the car would be 14.49 m/s
7 0
4 years ago
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